How do you find the equation of the tangent line to the curve #xy^2+e^y=2x^2-y+e^x-y+e^x# at (1,1)?

Question

Christopher Allers · Accepted Answer

Differentiate both sides to find the slope, and then use the slope and the point #(1,1)# to find the equation: #y=x# I think you added an extra #y# and #e^x# here by mistake, because the equation you provided does not contain the point #(1,1)#. However, #xy^2+e^y=2x^2-y+e^x# does contain #(1,1)#, so I'm assuming you meant that. We start by taking the derivative with respect to #x# on both sides - this is the first step in finding the slope of the tangent line: #d/dx(xy^2+e^y)=d/dx(2x^2-y+e^x#) Left Side We're trying to find #d/dx(xy^2+e^y)#. The sum rule says we can break this up into: #d/dx(xy^2)+d/dx(e^y)#. Starting with #d/dx(xy^2)#: #color(white)(XX)d/dx(xy^2)=(x)'(y^2)+(x)(y^2)'->#using product rule #color(white)(XX)d/dx(xy^2)=y^2+2xydy/dx# Onto #d/dx(e^y)#: #color(white)(XX)d/dx(e^y)=e^ydy/dx# Therefore #d/dx(xy^2+e^y)=y^2+2xydy/dx+e^ydy/dx#. That makes our problem come down to: #y^2+2xydy/dx+e^ydy/dx=d/dx(2x^2-y+e^x)# Right Side We're trying to find #d/dx(2x^2-y+e^x)#. Again, we can use the sum rule to break it down into: #d/dx(2x^2)-d/dx(y)+d/dx(e^x)#. These derivatives aren't too hard to manage: #color(white)(XX)d/dx(2x^2)=4x# #color(white)(XX)d/dx(y)=dy/dx# #color(white)(XX)d/dx(e^x)=e^x# Therefore #d/dx(2x^2-y+e^x)=4x-dy/dx+e^x#. That means our problem is: #y^2+2xydy/dx+e^ydy/dx=4x-dy/dx+e^x# Rearranging to solve for #dy/dx#, we see: #y^2+2xydy/dx+e^ydy/dx=4x-dy/dx+e^x# #2xydy/dx+e^ydy/dx+dy/dx=4x+e^x-y^2# #dy/dx(2xy+e^y+1)=4x+e^x-y^2# #dy/dx=(4x+e^x-y^2)/(2xy+e^y+1)# Therefore the slope of the tangent line is given by the expression: #(4x+e^x-y^2)/(2xy+e^y+1)#. We're trying to find the equation, which means finding the slope; and to do that, we simply find #dy/dx# at #(1,1)#: #dy/dx=(4x+e^x-y^2)/(2xy+e^y+1)# #dy/dx=(4(1)+e^((1))-(1)^2)/(2(1)(1)+e^((1))+1)# #dy/dx=(3+e^(1))/(3+e^(1))=1# Thus the slope of the tangent line is #1#. The equation will be of the form: #y=mx+b#, where #x# and #y# are points on the line, #m# is the slope, and #b# is the #y#-intercept. We know everything except #b#, but we have everything except #b#; that means we can use #x#, #y#, and #m# to find #b#: #y=mx+b# #1=(1)(1)+b# #0=b# That means the equation of the tangent line is #y=x#.

Evelyn Agard · Answer

undefined To find the equation of the tangent line to the curve at the point (1,1), we need to find the slope of the tangent line and then use the point-slope form of a line to write the equation.

1. Differentiate the equation implicitly with respect to x.
2. Substitute the coordinates of the point (1,1) into the derivative to find the slope.
3. Use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, to write the equation of the tangent line.

By following these steps, you can find the equation of the tangent line to the curve at (1,1).